What we should know about Linear Algebra

Transcription

1 Wht we should know out iner Alger D Coordinte geometr Vectors in spce nd spce Dot product nd cross product definitions nd uses Vector nd mtri nottion nd lger Properties (mtri ssocitiit ut NOT mtri commuttiit) Mtri trnspose nd inerse definition, use, nd clcultion Homogenous coordintes cs stff Timeless iner Alger /5

2 Shopping emple summr From Trnsformtions lecture Column ector for quntities, q: Row ector for corresponding prices t the stores (P): 6 5 store A (Est Side) store B (B & C) store C (Store 4) [ ] [ ] [ ] Sum: totlcost P( All) totlcost totlcost A B C The totlcost ector is determined row-column multipliction where row price, column quntities, i.e. dot product of price row with quntities column cs stff Timeless iner Alger /5

3 Clculting Cost for More thn One Buer For two uers, we need mtri-mtri multipliction. The new uer mtri will he two columns (one for ech uer) mtri 4 4 P( All) totlcost totlcost totlcost A B C totlcost totlcost totlcost A B C For n different uers, simpl etend the two uer sitution. Now, we he uer mtri which hs n columns cs stff Timeless iner Alger /5

4 Crtesin Coordinte Spce Emples: one, two nd three dimensionl rel coordinte spces R Y R Point t (, ) 4 5 Rel numers: etween n two rel numers on n is there eists nother rel numer Compre with the computer screen, positie integer coordinte sstem norml coordinte es X 4 5 cs stff Timeless iner Alger 4/5 Z R Y computer screen es Point t (4, -, ) 4

5 Vectors & Vector Spce (/) Consider ll loctions in reltionship to one centrl reference point, clled origin (, 4) 4 (8, 9) (, ) A ector tells ou which direction to go with respect to the origin, nd length of the trip A ector does not specif where the trip egins, to continue this nlog cs stff Timeless iner Alger 5/5

6 Vectors & Vector Spce (/) Vectors re used etensiel in computers to represent positions of ertices of ojects determine orienttion of surfce in spce ( surfce norml ) represent reltie distnces nd orienttions of lights, ojects, nd iewers in D scene, so the rendering lgorithms cn crete the impression of light intercting with solid nd trnsprent ojects (e.g., ectors from light sources to surfces) cs stff Timeless iner Alger 6/5

7 Vector ddition in Vector ddition R Fmilir ddition of rel numers [], [6], [9] Vector ddition in R The nd prts of ectors cn e dded using ddition of rel numers long ech of the es (component-wise ddition) ' 5 4 ' Result,, plotted in is the new ector R cs stff Timeless iner Alger 7/5

8 Adding Vectors Visull dded to, using the prllelogrm rule: tke ector from the origin to ; reposition it so tht its til is t the hed of ector ; define s the hed of the new ector or, equilentl, dd to cs stff Timeless iner Alger 8/5

9 cs stff Timeless iner Alger 9/5 On, fmilir multipliction rules On lso Sclr Multipliction (/) R R ( ) Y X 4 6

10 Sclr Multipliction (/) : iner Dependence Set of ll sclr multiples of ector is line through the origin Two ectors re linerl dependent when one is multiple of the other Y Definition of dependence for three or more ectors is trickier α X cs stff Timeless iner Alger /5

11 Bsis Vectors of the plne The unit ectors (i.e., whose length is one) on the nd -es re clled the stndrd sis ectors of the plne The collection of ll sclr multiples of gies the first coordinte is The collection of ll sclr multiples of gies the second coordinte is Then n ector cn e epressed s the sum of sclr multiples of the unit ectors: We cll these two ectors sis ectors for R ecuse n other ector cn e epressed in terms of them This is er importnt concept. Mke sure tht ou understnd it. cs stff Timeless iner Alger /5

12 cs stff Timeless iner Alger /5 nd re perpendiculr. Is this necessr? Questions rephrsed: cn we mke n ector Or, is there solution to the following, where nd re unknown: This is just liner sstem of two equtions. When is this unsolle? How does tht mke sense geometricll? The ectors cnnot e linerl dependent. Non-orthogonl Bsis Vectors m n from sclr multiples of rndom ectors nd d c d c d c m n β α β α β α α β OK OK not OK not OK

13 Uses of the dot product Define length or mgnitude of ector Normlize ectors (generte ectors whose length is, clled unit ectors) Mesure ngles etween ectors Determine if two ectors re perpendiculr Find the length of ector projected onto coordinte is There re pplets for the dot product under Applets -> iner Alger cs stff Timeless iner Alger /5

14 Rule for Dot product Also known s sclr product, or inner product. The result is sclr (i.e., numer, not ector) Defined s the sum of the pirwise multiplictions Emple: [ c d] cz dw z w Note, the result is not ector of the component-wise multiplictions, it is sclr lue Mn CPUs support the dot product of flots in MAC ector instruction: Multipl nd ccumulte, tht tkes two -D rrs s opernds (this instruction eists ecuse the dot product is computed so often in so mn different pplictions) cs stff Timeless iner Alger 4/5

15 Finding the ength of Vector ( ) The dot product of ector with itself,, is the squre of the length of the ector: We define the norm of ector (i.e., its length) to e Thus, ( ) for ll, with ( ) if nd onl if V is clled unit ector if, nd is denoted ˆ To mke n ritrr ector into unit ector, i.e. to normlize it, diide the length (norm) of, which is denoted. Note tht if, then its unit ector is undefined. So in generl (with the eception) we he: ˆ cs stff Timeless iner Alger 5/5

16 Finding the Angle Between Two Vectors The dot product of two non-zero ectors is the product of their lengths nd the cosine of the ngle etween them: ' ' cos( θ φ) sin(θ) rdius V Proof: [ ] [ cosθ sinθ ] [ ] ' [ cosφ sinφ ] cos(θ) cosθ cosφ cosθ cosφ sinθ sinφ sinθ sinφ cos θ cosφ sinθ sinϕ ( ) nd, sic trigonometric identit cosθ cosφ sinθ sinφ cos( θ φ ) so, cos( θ φ ) θ (θ -φ) φ V` cs stff Timeless iner Alger 6/5

17 More Uses of the Dot Product Finding the length of projection u If u is unit ector, then is the length of the projection of onto the line contining u V θ (θ -φ) φ u Perpendiculr ectors lws he dot product of ecuse the cosine of 9 o is Emple: nd. Determining right ngles ' ' 4 ( 4 ) ( 4) π cos cs stff Timeless iner Alger 7/5

18 cs stff Timeless iner Alger 8/5 Cross product of nd written is defined s: The resulting ector is perpendiculr to oth originl ectors. Tht is, it is norml to the plne contining the two ectors. Its length is equl to the re of the prllelogrm formed the two ectors Thus, we cn write: Note tht the cross-product does not generte normlized ector ( ector of unit length). An esier w to represent the mth for the cross product: (This is for those who know how to get the determinnt of mtri. i, j, k re the unit sis ectors.) Cross product follows right-hnd rule, so switching order of ectors gies ector in opposite direction. Cross Product, ˆ ˆ ˆ det k j i nd the ngle etween is where, sin θ θ ) ( ) in fct (,

19 Algeric Properties of Vectors Commuttie (ector) Associtie (ector) Additie identit Additie inerse Distriutie (ector) Distriutie (sclr) Associtie (sclr) Multiplictie identit ' ' ( ' ) u ( ' u) There is ector, such tht for ll, ( ) ( ) For n there is ector such tht ( ) r ( ' ) r r' ( r s) r s r ( s) ( rs) For n, R, cs stff Timeless iner Alger 9/5

20 Prolem: Trnslting Ojects Gol: moe the house from here to there R Y there (7, 6) here (, ) X Now tht we he ectors nd know how to mnipulte them, this isn t hrd We need to moe the house 5 in nd in 5 So just dd to ech erte cs stff Timeless iner Alger /5

21 Wht is Mtri? A mtri cn e esil isulized s ector of ectors. Specificll (n) rows of (m) dimensionl ectors. Intuitiel, it s D rr of numers For the purposes of this clss, mtrices re used to represent trnsformtions which ct on points, ectors, nd other mtrices et s quickl recp the tpes of trnsformtions we would like to e le to ccomplish cs stff Timeless iner Alger /5

22 Trnsltion Component-wise ddition of ectors ' d t where, ', t nd ' d d d To moe polgons: just trnslte ertices (ectors) nd then redrw lines etween them Preseres lengths (isometric) Preseres ngles (conforml) d d Y Note: House shifts position reltie to origin X NB: A trnsltion (,), i.e. no trnsltion t ll, gies us the identit mtri, s it should cs stff Timeless iner Alger /5

23 cs stff Timeless iner Alger /5 Component-wise sclr multipliction of ectors S where nd Does not presere lengths Does not presere ngles (ecept when scling is uniform) Scling ' ' ', s s S s s ' ' Note: House shifts position reltie to origin Y X s s NB: A scle of in oth directions, i.e. no scle t ll, gies us the identit mtri, s it should

24 Rottion Rottion of ectors through n ngle θ R θ where cosθ sinθ Rθ sinθ cosθ nd cos θ sin θ sin θ cos θ, ' ' ' Proof is doule ngle formul Confused? ook t sis ectors indiidull Preseres lengths nd ngles Y π θ 6 θ Note: house shifts position reltie to the origin NB: A rottion ngle, i.e. no rottion t ll, gies us the identit mtri, s it should X cs stff Timeless iner Alger 4/5

25 Digression: Tpes of Trnsformtions Projectie ffine liner iner: preseres prllel lines, cts on line to ield either nother line or point. The ector [, ] is lws trnsformed to [, ]. Emples: Scle nd Rotte. Affine: preseres prllel lines, cts on line to ield either nother line or point. The ector [, ] is not lws trnsformed to [, ]. Emples: Trnslte, s well s Rotte, nd Scle (since ll liner trnsformtions re lso ffine). Projectie: prllel lines not necessril presered, ut cts on line to ield either nother line or point (not cures). Emples: ou ll see trnsformtion in our cmer model which is not iner or Affine, ut is projectie. Trnslte, Rotte, nd Scle re ll Projectie, since this is een more generl thn Affine. All of these trnsformtions send lines to lines, therefore we onl need to store endpoints Note tht we re just tlking out trnsformtions here, nd not mtrices specificll (et) cs stff Timeless iner Alger 5/5

26 Sets of iner Equtions nd To trnslte, scle, nd rotte ectors we need function to gie new lue of, nd function to gie new lue of Emples For rottion ' cosθ sinθ For scling ' s ' s Mtrices ( ) ( sinθ cosθ ) ' These two, ut not trnsltion, re of the form ' ' c d A trnsformtion gien such sstem of liner equtions is clled liner trnsformtion nd is represented mtri: c d cs stff Timeless iner Alger 6/5

27 cs stff Timeless iner Alger 7/5 The new ector is the dot product of ech row of the mtri with the originl column ector. Thus, the kth entr of the trnsformed ector is the dot product of the kth row of the mtri with the originl ector. Emple: scling the ector 7 in the direction nd.5 in the direction Mtri-Vector Multipliction (/) d c M M d c d c,, where 6 ( ) ( ) ( ) ( )

28 cs stff Timeless iner Alger 8/5 In generl: Cn lso e epressed s: Mtri-Vector Multipliction (/) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) n n n n n n n n n n n n n m m m m c c c c m m m m c c c c m m MX n i i i n i i i n i i i

29 Algeric Properties of Mtrices Properties of mtri ddition Commuttie Associtie Identit Inerse Distriutie (mtri) Distriutie (sclr) Properties of mtri multipliction NOT commuttie Associtie (mtri) Associtie (sclr) Distriutie (ector) Identit Inerse (more lter) A B B A A B C A B C ( ) ( ) There is mtri, such tht, for ll A, A A A ( ) ( ) For n A, there is A such tht A ( A) r ( A B) ra rb r s A ra ( ) sa AB BA ( AB ) C A( BC) ( rs ) A r( sa) A ( ' ) A A' There is mtri I, such tht, for ll A, AI A IA For some A, there is mtri A - such tht: AA I A A cs stff Timeless iner Alger 9/5

30 Mtri Multipliction, isull Now tht we know tht mtri multipliction is distriutie oer ectors, we cn understnd liner trnsformtions etter Suppose we he some mtri, like, nd we wnt to know wht it will do to ojects. First, multipl the two unit sis ectors: Notice tht the results re the two columns of the mtri. Now, since we sid tht mtri multipliction is distriutie oer ectors, we cn tell where n ector will e sent knowing where ech unit sis ector is sent So if we he some trnsformtion mtri, isulize wht it does to n oject cs stff Timeless iner Alger /5

31 Visul Mtri Multipliction, continued (/) Originll, we he unit grid: Then we ppl the trnsformtion nd set up grid sed on the imges of the two unit/sis ectors: cs stff Timeless iner Alger /5

32 Visul Mtri Multipliction, continued (/) Now, to see where ends up, insted of going units in nd in, go steps long the imge of nd long the imge of We cn see wht this looks like in the originl coordinte sstem: This emple effectiel illustrtes the chnge of sis concept. The row ectors of the mtri define the new sis. cs stff Timeless iner Alger /5

33 Mtri-Mtri Multipliction (/) Generll, how do we compute the product MN from mtrices M nd N? One w to think of mtri multipliction is in terms of row nd column ectors: MN ij the dot product of the ith row of M nd the jth column of N It is importnt to note tht if M is n m n mtri, then N must e n n k mtri. This is ecuse the rows of M nd the columns of N must e the sme size in order to compute their dot product So for M, n m n mtri, nd N, n n k mtri: MN row row M, M, m coln M M coln,, row M, i coln O, j O O row row M, m M, m coln M M coln, k, k Where row A, mens the th row of A, nd similrl col A, mens the th column of A cs stff Timeless iner Alger /5

34 cs stff Timeless iner Alger 4/5 Mtri-Vector Multipliction (/) Sometimes it cn help lot to he picture of this process in our hed For emple, let s s tht N is n n mtri. Then MN cn e clculted in the following mnner. Γ Γ Γ Γ Γ n n n n n M M β β α α β β β α α α M N MN

35 Comining Trnsformtions Mtrices representing trnsformtions performed in sequence cn e composed into single mtri Howeer, there re prolems with comining mtrices Mtri multipliction is not commuttie. Rotting or scling oject not centered t the origin introduces unwnted trnsltion Trnsltion induced scling nd rottion Rottion followed non-uniform scle; net sme scle followed sme rottion cs stff Timeless iner Alger 5/5

36 cs stff Timeless iner Alger 6/5 Mtri Composition Rule for composing mtrices The product of mtrices M M is thus creted from the dot products of the rows of M with the columns of M h g f e M d c M, For nd ) )( ( )) ( ( M M M M ) ( ) ( ( ) ( h g d f e c h g f e h g f e d c h g f e d c dh cf dg ce h f g e dh cf dg ce h f g e ) )( ( ) )( ( ) ( ) ( ) ( ) ( ' h f d c g e d c h f g e M M

37 Commuttie nd Non- Commuttie Comintions of Trnsformtions Be Creful! In D Commuttie Trnslte, trnslte Scle, scle Rotte, rotte Scle uniforml, rotte Non-commuttie Non-uniform scle, rotte Trnslte, scle Rotte, trnslte In D Commuttie Trnslte, trnslte Scle, scle Scle uniforml, rotte Non-commuttie Non-uniform scle, rotte Trnslte, scle Rotte, trnslte Rotte, rotte cs stff Timeless iner Alger 7/5

38 Trnsltion reisited et s look ck t the house trnsltion prolem: R Y there 6 (7, 6) 5 4 here (, ) We decided to do this mthemticll, dding ector to n erte in the scene. ' d ' d et s rewrite the mth it ' * * d* ' * * d * * * * Hmmm, this looks fmilir ' d ' d This m show ou more logicll how we decided to use homogenous coordintes. It s huge gin, ecuse we cn now compose trnsltions long with other trnsformtions in uniform mnner. or ' ' X d d * cs stff Timeless iner Alger 8/5

39 Mtri Inerse In the iewing lecture we will he use for M -, so here we ll tlk etensiel out wht the inerse does. Also, how to find M -, gien M (if it eists) Recll the definition: AA I A A The rule for inerting composed mtrices is: ( AB) B A It is importnt to note tht mtri is not lws inertile. A mtri will not e inertile if: It is not squre (n n mtri) It hs row / column of ll zeros An row / column is multiple of n other row / column (clled linerl dependent) Mtrices for rottion, trnsltion (using homogeneous coordintes), nd scling will lws he inerses. This mkes us hpp cs stff Timeless iner Alger 9/5

40 Wht Does n Inerse Do? The inerse A - of trnsformtion A will undo the result of trnsforming A For emple: if A scles fctor of two nd rottes 5 o, then A - will rotte 5 o nd then scle one hlf Rectngle(B) A*B A - *B A - *A*B cs stff Timeless iner Alger 4/5

41 Wht ou re out to see Suppose we he some mtri A tht we re tring to inert. et s write n oious eqution, A I * A Now, suppose we multipl ech side some mtrices, one one, i.e. T *T *A T *T *I*A And suppose fter we e multiplied enough of these, we end up with the identit on the left side, so we he I T n * * T *I*A Then on the left we hd T n * * T *A, so T n * * T A -, the definition of A - But now, on the right, (T * *T *I)A -, so s long s we didn t multipl A et, we he isolted A - So, to sum up: Strt with oth A nd I Perform the sme opertions on ech, until A ecomes I. Now, where we strted with I, we re left with A - cs stff Timeless iner Alger 4/5

42 How to Inert Mtri (/4) We re going to use Guss-Jordn elimintion Finding A - with Guss-Jordn elimintion is done ugmenting A with I to get [A I], then reducing the new mtri into reduced row echelon form (rref) to get new mtri. This new mtri will e of the form [I A - ] Wht does rref rell men? It mens the mtri hs the following properties: If row does not consist entirel of zeros, then the first nonzero numer in the row is. (Cll this leding ) If there re n rows tht consist entirel of zeros, then the re grouped together t the ottom of the mtri. If n two successie rows do not consist entirel of zeros, the leding in the lower row occurs frther to the right thn the leding in the higher row Ech column tht contins leding hs zeros eerwhere else. cs stff Timeless iner Alger 4/5

43 How to Inert Mtri (/4) To trnsform mtri into rref we re llowed to perform n of the three elementr row opertions. These re: Multipl row nonzero constnt Interchnge two rows Add multiple of one row to nother row Notice tht these cn ll e represented s mtrices. So wht we re rell doing is uilding up the inerse it it, nd multipling the mtrices the Identit s well s our originl mtri just so we will still he the inerse once we re done uilding it up. For emple, the following mtri cn e used to multipl the first row 4 in mtri: 4 This mtri swps the first rows in mtri: cs stff Timeless iner Alger 4/5

44 How to Inert Mtri (/4) Note tht with these opertions ou cn: Crete leding diiding the leding constnt in row: (row op ) Moe zero rows to the ottom of the mtri interchnging rows: (row op ) Assure tht zeros occur s ll entries, ecept the leding, in column cnceling out the other lues with multiples of : (row op ) It is importnt to note tht we would he no use for the mtri t n intermedite stge. Although we re ppling strictl liner trnsformtions, the he no geometric mening, the re just w of chieing the end result it it. cs stff Timeless iner Alger 44/5

45 How to Inert Mtri (4/4) (note: these steps tken from Elementr iner Alger, Howrd Anton, which cn e found in the Scii) The steps we will ctull use: Step. octe the leftmost column tht does not consist entirel of zeros. Step. Interchnge the top row with nother row, if necessr, to ring nonzero entr to the top of the column found in Step. Step. If the entr tht is now t the top of the column found in Step is, multipl the first row / in order to introduce leding. Step 4. Add suitle multiples of the top row to the rows elow so tht ll entries elow the leding ecome zeros. Step 5. Now coer the top row in the mtri nd egin gin with Step pplied to the sumtri tht remins. Continue in this w until the entire mtri is in row-echelon form. (not reduced rowechelon form) Step 6. Beginning with the lst nonzero row nd working upwrd, dd suitle multiples of ech row to the rows oe to introduce zeros oe the leding s. cs stff Timeless iner Alger 45/5

46 Inersion Emple (/) This will e smll emple ecuse doing this hnd tkes while. et s find A - for: A 7 9 Augment this with the identit to get [ A I] 7 9 Row opertion, multipl row /. 7 9 Row opertion, multipl row 7 nd dd it to row 7 cs stff Timeless iner Alger 46/5

47 Inersion Emple (/) Row opertion, multipl row / 7 Row opertion, multipl row / nd dd to row [ I A ] 9 7 Therefore: A 9 7 cs stff Timeless iner Alger 47/5

48 cs stff Timeless iner Alger 48/5 For more detil, refer to the inersion hndout or Anton s ook. The generl pttern: On the left, incrementll go from A to I to to Inersion Emple (/) * * * * * * * * * * * O M M O M M * * * * O M M (ll positions ritrr numers - A) (ones long digonl, zeroes elow, ritrr numers oe) (ones long digonl, zeroes elow nd oe - Identit)

49 cs stff Timeless iner Alger 49/5 The ppliction of mtrices in the row ector nottion is eecuted in the reerse order of pplictions in the column ector nottion: Column formt: ector follows trnsformtion mtri. Row formt: ector precedes mtri nd is posmultiplied it. B conention, we lws use column ectors. Addendum Mtri Nottion z i h g f e d c z ' ' ' [ ] [ ] i h g f e d c z z ' ' ' [ ] z z

50 But, There s Prolem Notice tht g c h i [ z] d e f [ d gz e hz c f iz] while d g e h c cz f d e fz i z g h iz cs stff Timeless iner Alger 5/5

51 Solution to Nottionl Prolem In order for oth tpes of nottions to ield the sme result, mtri in the row sstem must e the trnspose of the mtri in the column sstem Trnspose is defined such tht ech entr t (i,j) in M, is mpped to (j,i) in its trnspose (which is denoted M T ). You cn isulize M T s rotting M round its min digonl M d g e h c f, M i T c d e f g h i Agin, the two tpes of nottion re equilent: d g c f i cz d e fz d e g h iz g h [ z] e h [ cz d e fz g h iz] c f i z Different tets nd grphics pckges use different nottions. Be creful! cs stff Timeless iner Alger 5/5

52 Mtri Nottion nd Composition Appliction of mtrices in row nottion is reerse of ppliction in column nottion: TRS Mtrices pplied right to left S T R T T T Mtrices pplied left to right cs stff Timeless iner Alger 5/5

53 Summr We cn do sic mth with ectors nd mtrices We know how to compute the dot product of two ectors, nd rious uses for the dot product (nd don t worr, ou ll e using it for ll these things) We e soled the trnsltion prolem We cn compose mn mtrices so tht we onl he to do one mtri-ector multipliction to trnsform n oject ritrril We cn inert mtri (for more detil, refer to Elementr iner Alger, Howrd Anton, which cn e found in the Sci i) cs stff Timeless iner Alger 5/5